Maximizing edges

[Thanic Nur Samin]

Let there be $n$ points in a space. Some edges are connecting them, making a graph. Maximize the number of edges so that there is no tetrahedron in the graph.

[Thanic Nur Samin]

We claim that the answer is $\lfloor \dfrac{n^2}{3} \rfloor$. We induct on $n$.
The fact is obvious when $n=1,2,3$. If $n>3$, then, assume that it is true for $n=m$. Now, for $n=m+3$, consider vertices, connected with each other. If no such triangle exists, then the number of edges are less than $\dfrac{(m+3)^2}{3}$ [This is fairly easy with extremal principal]. Now, remove the $3$ vertices. The graph still doesn't contain any tetrahedron. Besides the $3$ edges connecting these vertices, we determine what is the most number of edges that could be there but was erased upon the removal of the $3$ vertices. Now, there are $m$ vertices left. If any vertex was connected to all three of them, then there would be a tetrahedron, which leads to a contradiction. So any vertex would be connected to at most $2$ of them, leading to a maximum number of $2m$ vertices removed. So altogether, at most $2m+3$ vertices are removed. The construction of the case of $2m+3$ vertices removed can be done easily by neglecting on vertex of the triangle and connecting the other two to all the other vertexes. So, the number of maximum edges for $n=m+3$ is $\le \dfrac{m^2}{3}+2m+3=\dfrac{(m+3)^2}{3}$, and so the maximum number of edges is $\lfloor \dfrac{n^2}{3} \rfloor$, by induction.

[asif e elahi]

Let there be $n$ points in a space. Some edges are connecting them, making a graph. Maximize the number of edges so that there is no tetrahedron in the graph.

This is just a special case of Turan's theorem.
https://en.m.wikipedia.org/wiki/Tur%C3%A1n's_theorem